Given $A \subseteq \mathbf{R}$, let $L$ be the set of all limit points of $A$. One can show that the set $L$ is closed. Now, argue that if $x$ is a limit point of $A \cup L$, then $x$ is a limit point of $A$. This can then be used to prove that $\overline{A} = A \cup L$ (the closure of $A$) is closed, i.e., contains its limit points.
The proof uses the fact that $x$ is either a limit point of $A$ or $L$. How do we know this? I understand that $x$ is the limit of a sequence $(x_n)$ with terms in $A \cup L$, but how does this imply that the sequence is either in $A$ or in $L$? Why can't the sequence have some points in $A$ and some in $L$? Besides this, the argument is straightforward.
Best Answer
Hint: ${\bf R}$ is a metric space, so any limit point is the limit of some sequence. If $x_n$ is a sequence of points in $L\cup A$, then either for infinitely many $n$ we have $x_n\in A$ or for infinitely many $n$ we have $x_n\in L$.